Particular Solution

Using substitution $$u = x^2 + y^2$$:
$$\frac{du}{dx} = 2x + 2y\frac{dy}{dx}$$

$$y\frac{dy}{dx} = \frac{1}{2}\frac{du}{dx} - x$$

Substituting into the original equation:
$$\frac{1}{2}\frac{du}{dx} - x + x = \sqrt{u}$$

$$\frac{1}{2}\frac{du}{dx} = \sqrt{u}$$

$$\frac{du}{dx} = 2\sqrt{u}$$

Separating variables:
$$\frac{du}{\sqrt{u}} = 2dx$$

$$2\sqrt{u} = 2x + C$$

$$\sqrt{u} = x + \frac{C}{2}$$

Applying the initial condition at $$x = 0, y = 1$$: $$u = 1$$
$$\sqrt{1} = 0 + \frac{C}{2}$$
$$C = 2$$

Therefore:
$$\sqrt{x^2 + y^2} = x + 1$$

$$x^2 + y^2 = (x+1)^2 = x^2 + 2x + 1$$

$$y^2 = 2x + 1$$